# `f(x) = cot(x/2), [pi,3pi]` Explain why Rolle’s Theorem does not apply to the function even though there exist `a` and `b` such that `f(a) = f(b)`.

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The Rolle's theorem is applicable to the given function, only if the function is continuous and differentiable over the interval, and f(a) = f(b). Since all trigonometric functions are continuous and differentiable on R, hence, the given function is continuous and differentiable on interval. Now, you need to check if `f(pi) = f(3pi).`

`f(pi) = cot (pi/2) = 0`

`f(3pi) = cot(3pi/2) = 0`

Since all the three conditions are valid, you may apply Rolle's theorem:

` f'(c)(b-a) = 0`

Replacing `3pi` for b and `pi` for a, yields:

`f'(c)(3pi-pi) = 0`

You need to evaluate f'(c), using chain rule:

`f'(c) = (cot(c/2))' = -1/(2sin^2(c/2)) `

Replacing the found values in equation `f'(c)(3pi-pi) = 0.`

`-2pi/(2sin^2(c/2)) != 0 `

**Hence, in this case, there is no valid value of c for Rolle's theorem to be applied.**